# Actual double slit

I can't understand what is given here. How is the double slit pattern a interference a two single slit and double slit. If we say it is interference of two single slit then it makes sense. But the graph looks different from double slit pattern. Please explain

• Aug 2 '18 at 20:29

The "double-slit interference pattern," as defined in this passage, is what you get when you shine light on two slits of infinitesimal width. In reality, you can't ever make slits of infinitesimal width, so whenever you actually perform this experiment, you must account for the fact that your slits have finite width. Shining light through a slit of finite width gives a single-slit interference pattern. So the "actual double-slit interference pattern" (i.e. the pattern you will get in every double-slit experiment you could possibly perform) is a combination of the "double-slit interference pattern" (the ideal one, with infinitesimal slit widths) and two single-slit interference patterns.

• So since the widths are not Infinitesimal we are getting the effects of the single slit pattern Oct 15 '18 at 6:50

Reading the words carefully, I think the text is right.

This is the diffraction pattern from the left slit:

This is the diffraction pattern from the right slit:

right diffraction pattern

And this is the superposition of both diffraction patterns, just the sum of the previous two:

Up to here it is the classical expected behevior.

But then there is the double-slit interference pattern:

Combine these two last images (multiply them, I think) and you'll get the one from your book.

The last paragraph about $a$ being much smaller than $d$ is so that the interference pattern is much narrower than the diffraction pattern. If they are of the same magnitude (or $a$ greater than $d$) then the quantum effect is not clearly visible, which is the point of this experiment.